/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 60 Identify the types of hybrid orb... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 60

Identify the types of hybrid orbitals on the central atom that form the \(\sigma\) bonds in the following molecules. (a) \(\mathrm{ClF}_{3}\) (b) \(\mathrm{BBr}_{3}\) (c) \(\mathrm{BeF}_{2}\) (d) \(\mathrm{ONCl}\)

Short Answer

Expert verified
(a) sp鲁d; (b) sp虏; (c) sp; (d) sp虏

Step by step solution

01

Determine the Central Atom

For each molecule, identify the central atom that is bonded to other atoms. (a) In \(\mathrm{ClF}_{3}\), the central atom is \(\text{Cl}\).(b) In \(\mathrm{BBr}_{3}\), the central atom is \(\text{B}\).(c) In \(\mathrm{BeF}_{2}\), the central atom is \(\text{Be}\).(d) In \(\mathrm{ONCl}\), the central atom is \(\text{N}\).
02

Count the Bond Pairs and Lone Pairs

Find the number of sigma bonds and lone pairs on the central atom to determine hybridization.(a) \(\mathrm{ClF}_{3}\): 3 sigma bonds and 2 lone pairs.(b) \(\mathrm{BBr}_{3}\): 3 sigma bonds and 0 lone pairs.(c) \(\mathrm{BeF}_{2}\): 2 sigma bonds and 0 lone pairs.(d) \(\mathrm{ONCl}\): 2 sigma bonds and 1 lone pair.
03

Determine Hybridization Based on Bonds and Lone Pairs

Use the information from Step 2 to determine the hybridization of each central atom.Hybridization can be deduced using the formula: Number of sigma bonds + Number of lone pairs = Total pairs- 2 pairs: \(\text{sp}\)- 3 pairs: \(\text{sp}^2\)- 4 pairs: \(\text{sp}^3\)- 5 pairs: \(\text{sp}^3d\)(a) \(\mathrm{ClF}_{3}\): 5 total pairs = \(\text{sp}^3d\) hybridization.(b) \(\mathrm{BBr}_{3}\): 3 total pairs = \(\text{sp}^2\) hybridization.(c) \(\mathrm{BeF}_{2}\): 2 total pairs = \(\text{sp}\) hybridization.(d) \(\mathrm{ONCl}\): 3 total pairs = \(\text{sp}^2\) hybridization.
04

Identify the Hybrid Orbitals Forming Sigma Bonds

Based on the hybridization calculated, identify which hybrid orbitals form the sigma bonds in each molecule.(a) \(\mathrm{ClF}_{3}\): \(\text{sp}^3d\) hybrid orbitals form sigma bonds.(b) \(\mathrm{BBr}_{3}\): \(\text{sp}^2\) hybrid orbitals form sigma bonds.(c) \(\mathrm{BeF}_{2}\): \(\text{sp}\) hybrid orbitals form sigma bonds.(d) \(\mathrm{ONCl}\): \(\text{sp}^2\) hybrid orbitals form sigma bonds.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sigma Bonds
Sigma bonds (\( \sigma \) bonds) are a type of covalent bond that form between atoms. They are the strongest type of covalent bond and result from the direct overlap of atomic orbitals. In a sigma bond, the overlapping occurs along the axis connecting the two nuclei of the bonding atoms, forming a cylindrical symmetry around this axis. This makes sigma bonds robust and integral in holding atoms together within a molecule.
  • They often occur alongside pi bonds in double and triple bonds, but sigma bonds are always the first bond to form.
  • Sigma bonds are crucial in determining the molecular structure and strength of a bond.
  • In molecules like \( \text{ClF}_{3} \), \( \text{BBr}_{3} \), and \( \text{BeF}_{2} \), the sigma bonds are formed by the hybrid orbitals of the central atom.
Understanding sigma bonds is a foundational aspect of studying molecular geometry and hybridization, as they play a key role in the way molecules are formed and how they behave in various chemical reactions.
Lone Pairs
Lone pairs of electrons are the valence electrons that are not shared with another atom and do not participate in bonding. In contrast to sigma bonds, lone pairs reside on a single atom and can influence the shape and geometry of a molecule significantly.
  • Lone pairs can occupy space around a central atom and can repel other electron pairs due to their charge, affecting the overall molecular geometry.
  • They are included in hybridization calculations because they occupy orbitals around the central atom, impacting the total number of orbitals needed for bonding.
  • For example, in \( \text{ClF}_{3} \), the presence of two lone pairs on the chlorine atom results in more complex molecular geometry.
The effects of lone pairs are crucial for understanding the 3D shape of molecules and help explain deviations in bond angles and molecular functionality.
Central Atom
The central atom in a molecule is typically the one that is bonded to two or more other atoms and acts as a foundation for molecular structure. It often plays a key role in determining the hybridization state of the molecule.
  • The central atom is chosen based on factors like electronegativity and atom size; it is usually the least electronegative atom that isn't hydrogen.
  • In \( \text{ClF}_{3} \), \( \text{BBr}_{3} \), and \( \text{BeF}_{2} \), the central atoms are Cl, B, and Be, respectively.
  • The properties and hybridization of the central atom largely determine the molecule鈥檚 ability to form structures and bond angles.
Identifying the central atom is a necessary step in understanding how a molecule is arranged and how it might interact in chemical reactions.
Molecular Geometry
Molecular geometry refers to the three-dimensional arrangement of atoms in a molecule. Understanding molecular geometry is crucial for predicting the behavior and reactivity of molecules.
  • It is influenced by both sigma bonds and lone pairs, as these determine the spatial orientation of atoms.
  • The VSEPR (Valence Shell Electron Pair Repulsion) theory is often used to determine geometry based on the repulsion between electron pairs.
  • For instance, \( \text{ClF}_{3} \) has a T-shaped geometry due to the two lone pairs pushing the sigma bonds into this configuration.
Gaining insight into molecular geometry is vital for predicting physical and chemical properties, offering a clearer picture of how molecules will interact in biological and chemical systems.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The reaction of sulfur, \(\mathrm{S}_{8}\), with fluorine, \(\mathrm{F}_{2}\), yields a product with the general formula \(\mathrm{SF}_{x}\). If \(4.01 \mathrm{~g} \mathrm{~S}_{8}\) reacts with \(4.76 \mathrm{~g} \mathrm{~F}_{2}\) to yield only \(\mathrm{SF}_{x},\) what is the value of \(x ?\) Draw the Lewis structure of this compound, indicating the \(\mathrm{F}-\mathrm{S}-\mathrm{F}\) bond angles and the hybrid orbitals on sulfur.

Use the VSEPR model to predict the bond angles around each central atom in the following Lewis structures (benzene rings are frequently pictured as hexagons, without the letter for the carbon atom at each vertex). Note that the drawings do not necessarily depict the bond angles correctly.

Give the bonded-atom lon猫-pair arrangement and the molecular shape of the following species. (a) \(\mathrm{SeO}_{2}\) (b) \(\mathrm{N}_{2} \mathrm{O}(\mathrm{N}\) is the central atom \()\) (c) \(\mathrm{H}_{3} \mathrm{O}^{+}\) (d) \(\mathrm{IF}_{5}\) (e) \(\mathrm{SCl}_{4}\)

Phosgene, \(\mathrm{COCl}_{2}\), is a highly toxic gas that was used in combat during World War I. It is an important intermediate in the preparation of a number of organic compounds but must be handled with extreme care. Given that carbon is the central atom in phosgene, determine the Lewis structure, the bonded-atom lone-pair arrangement, the hybridization of the carbon atom, and the polarity of the molecule.

A Recently, the structure of an amine compound, \(\mathrm{NR}_{3}\) \((\mathrm{R}=\) large organic group), has been determined to have C-N-C bond angles of 119.2 degrees. It is believed that the bond angles of about 109 degrees expected from the VSEPR model are not observed because of the large substituents bonded to the nitrogen atom. Given this large bond angle, what type of orbital on the nitrogen atom makes the \(\mathrm{N}-\mathrm{C} \sigma\) bonds, and in what type of orbital is the lone pair located?
See all solutions

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Read Explanation

Chemical Analysis

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemical Reactions

Read Explanation

Inorganic Chemistry

Read Explanation

Physical Chemistry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.